Question

Verify that the force field is conservative. Then find a scalar potential φ such that$\mathbf{F}\mathbf{=}\mathbf{-}\mathbf{\nabla }\mathbf{\phi }\mathbf{,}\mathbf{F}\mathbf{=}\mathbf{yi}\mathbf{+}\mathbf{xj}\mathbf{+}\mathbf{k}$ ,

K = constant.

Short answer

The force field is conservative and the scalar potential is $-xy-z$

Step-by-step solution

Given Information

The force field is $\mathrm{F}=\mathrm{yi}+\mathrm{xj}+\mathrm{k}$

and $\mathrm{F}=-\nabla \mathrm{\phi }.$

Definition of conservative force and scalar potential.

A force is said to be conservative if $\mathbf{\nabla }\mathbf{\times }\mathbf{F}\mathbf{=}\mathbf{0}$.

The formula for the scalar potential is $\mathbf{W}\mathbf{=}\mathbf{\int }\mathbf{F}\mathbf{.}\mathbf{dr}\mathbf{.}$

Verify whether the force is conservative or not.

The force is said to be conservative if $\nabla \times F=0$.

Putthe values given below in the above equation.

$F=yi+xj+k$

The equation becomes as follows.

$$\begin{gathered} \nabla \times F=\left(\begin{array}{ccc}i& j& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ y& x& 1\end{array}\right) \\ \nabla \times F=\left(\frac{\partial }{\partial y}\left(1\right)-\frac{\partial }{\partial z}\left(x\right)\right)i-\left(\frac{\partial }{\partial x}\left(1\right)-\frac{\partial }{\partial z}\left(y\right)\right)j+\left(\frac{\partial }{\partial x}\left(y\right)-\frac{\partial }{\partial y}\left(x\right)\right)k \\ \nabla \times F=0 \end{gathered}$$

The field is conservative.

Define a formula for scalar potential.

The formula for the scalar potential is$\mathrm{W}=\int \mathrm{F}.\mathrm{dr}.$

$$\begin{gathered} \mathrm{W}=\int (\mathrm{yi}+\mathrm{xj}+\mathrm{k}).(\mathrm{dxi}+\mathrm{dyj}+\mathrm{dzk}) \\ \mathrm{W}=\int \mathrm{ydx}+\mathrm{xdy}+\mathrm{dz} \end{gathered}$$

Take the path from  (0,0,0)  to (x,y,z) and evaluate W.

$W_1$is from$(0,0,0)\mathrm{to}(\mathrm{x},0,0).$ .

$$\begin{gathered} y=0 \\ dy=0 \\ z=0 \\ dz=0 \end{gathered}$$

Substitute the above value in the equation mentioned below.

$$\begin{gathered} W=\int ydx+xdy+dz \\ {W}_1={\int }_0^{x}0\times dx \\ {W}_1=0 \end{gathered}$$

$W_2$is from $(\mathrm{x},0,0)\mathrm{to}(\mathrm{x},0,\mathrm{z})$x is constant.

$\begin{array}{l}dx=0\\ y=0\\ dy=0\end{array}$

Substitute the above value in the equation mentioned below.

role="math" localid="1659336578414" $$\begin{gathered} W=\int ydx+xdy+dz \\ {W}_2={\int }_0^{z}dz \\ {W}_2={\left[z\right]}_0^z \\ {W}_2=z \end{gathered}$$

$W_3$is from $(x,0,z)to(x,y,z)$and x is constant, z is constant.

$\begin{array}{l}dx=0\\ dz=0\end{array}$

Substitute the above value in the equation mentioned below.

role="math" localid="1659336294517" $$\begin{gathered} W=\int ydx+xdy+dz \\ {W}_3={\int }_0^{y}xdy \\ {W}_3={\left[xy\right]}_0^y \\ {W}_3=xy \end{gathered}$$

The formula states the equation mentioned below.

$\begin{array}{l}W=W_1+W_2+W_3\\ W=0+z+xy\\ W=Z+xy\end{array}$

Find the value of φ

The formula states the equation mentioned below.

$F=\nabla W$

It is given that$F=-\nabla \phi .$.

By both the values of , $\mathrm{F},-\nabla \mathrm{\phi }=\nabla \mathrm{W}.$

$\mathrm{\phi }=-\mathrm{W}$

Put the value of in above equation.

$\begin{array}{l}\phi =-\left(z+xy\right)\\ \phi =-z-xy\end{array}$

Hence the scalar potential is $-xy-z$